For the system of linear equations $αx + y + z = 1, x + αy + z = 1, x + y + αz = β$ , which one of the following statements is NOT correct ?

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It has infinitely many solutions if $α = 2 and β = - 1$

It has no solution if $α = - 2 and β = 1$

It has infinitely many solutions of $α = 1 and β = 1$

$x + y + z = \frac{3}{4} if α = 2 and β = 1$

answer is A.

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Detailed Solution

$Δ = |\begin{array}{lll} α 1 1 \\ 1 α 1 \\ 1 1 α \end{array}|$ $= (α - 1)^{2} (α + 2)$
$Δ_{1} = |\begin{array}{lll} 1 1 1 \\ 1 α 1 \\ β 1 α \end{array}|$ $= (α - β) (α - 1)$
$Δ_{2} = |\begin{array}{lll} α 1 1 \\ 1 1 1 \\ 1 β α \end{array}|$ $= (α - 1) (α - β)$
$Δ_{3} = |\begin{array}{lll} α 1 1 \\ 1 α 1 \\ 1 1 β \end{array}|$ $= (α - 1) (α β + β - 2)$
i) If $α = 2 & β = - 1$
$Δ \neq 0 \Rightarrow$ unique solution
ii) If $α = - 2 & β = 1$
$\Rightarrow Δ = 0, Δ_{1} \neq 0 \Rightarrow$ No solution
ii) If $α = 1; β = 1$
$\Rightarrow Δ = 0, Δ_{1} = Δ_{2} = Δ_{3} = 0$
$x + y + z = 1 \to$ coincidental planes
$\Rightarrow$ Infinite solutions
iv) $α = 2; β = 1 \Rightarrow Δ \neq 0$
$2 x + y + z = 1$
$\begin{aligned} x + 2 y + z = 1 \Rightarrow x = y = z = \frac{1}{4} \\ x + y + 2 z = 1 ∴ x + y + z = \frac{3}{4} \end{aligned}$